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In this page there are calculations of variation of Einstein-Hilbert action.

I see variations of terms like this:

$\delta {R^{\rho }}_{{\sigma \mu \nu }}$

where the term is not a functional, and

$\frac {\delta {\mathcal {L}}_{{\mathrm {M}}}}{\delta g^{{\mu \nu }}}$

where we have a functional derivative of a term that is not also a functional.

What is the exact meaning of those expressions?

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  • $\begingroup$ Not sure, but I think the variations are taken wrt the inverse metric (look at the part with the action). You can view $R$ and $\Gamma$ as functionals of $g$ (or here of $g^{-1}$). I think the notation they switch to later is more "physicsy", in that $\delta R$ means a perturbation of $R$ with an infinitesimal test function, so later on when they "divide" by $\delta g^{ij}$ it becomes the first variation wrt $g^{ij}$. $\endgroup$ – user3658307 Mar 27 '18 at 17:56
  • $\begingroup$ You can see here an answer: mathoverflow.net/questions/295599/… $\endgroup$ – asv Mar 28 '18 at 12:45
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The symbol "$\delta$" in calculus of variation, especially in physics, denotes the change of an output variable due to the change of input variable(s). In general, let $g:Y_1\times Y_2\times\cdots\times Y_n\to Z$ be a mapping, then $$ \left(\delta g\right)(y_1,y_2,\cdots,y_n):=g(y_1+\delta y_1,y_2+\delta y_2,\cdots,y_n+\delta y_n)-g(y_1,y_2,\cdots,y_n). $$

Here are some examples.

(1) Consider $f:\mathbb{R}\to\mathbb{R}$, which is a function. Then $$ \left(\delta f\right)(x)=f(x+\delta x)-f(x)=f'(x)\delta x+o(\delta x), $$ which is exactly the differentiation of this function, i.e., $$ \left({\rm d}f\right)(x)=f(x+{\rm d}x)-f(x)=f'(x){\rm d}x+o({\rm d}x). $$ It is conventional to keep only the first-order variation and drop the higher-order terms. Thus $$ \left(\delta f\right)(x)=f'(x)\delta x. $$

(2) Consider $J:\mathscr{F}\to\mathbb{R}$ with $\mathscr{F}$ being some function space, which is a functional. Then $$ \left(\delta J\right)(f)=J(f+\delta f)-J(f). $$

(3) Consider $g=\det\left(g_{\mu\nu}\right)$. Recall the definition of the determinant, and $g$ can be regarded as a polynomial function of all $g_{\mu\nu}$'s. For example, $$ \left(g_{\mu\nu}\right)=\left( \begin{array}{cc} a&b\\ c&d \end{array} \right)\Longrightarrow\det g=ad-bc, $$ which is a second-order multivariate polynomial. In this sense, $g$ is a function of all of its entries, and $\delta g$ is no more than the differentiation of $g$ with respect to those entries, i.e., $$ {\rm d}g=g^{\nu\mu}g{\rm d}g_{\mu\nu}\iff\delta g=g^{\nu\mu}g\delta g_{\mu\nu}. $$

(4) Consider $\mathcal{L}[x(t)]=\left(m/2\right)\dot{x}^2(t)-\Phi_g(x(t))$ with $\Phi_g$ being, say, gravitational potential, which is the integrant of the Newtonian action. In this case, "$\delta$" is in the sense of a shorthand of the respective functional derivative. That is, define $$ J[x(t)]=\int_{t_1}^{t_2}\mathcal{L}[x(t)]{\rm d}t, $$ and we have, using integration by parts, $$ \left(\delta J\right)[x(t)]=\int_{t_1}^{t_2}\left(-m\ddot{x}(t)-\left(\nabla\Phi_g\right)(x(t))\right)\delta x(t){\rm d}t. $$ With this in mind, denote $$ \delta\mathcal{L}[x]=\left(-m\ddot{x}-\left(\nabla\Phi_g\right)(x)\right)\delta x. $$ Remark. In this last example, $\mathcal{L}[\cdot]$ and $\mathcal{L}(\cdot)$ do not mean the same. $\mathcal{L}[\cdot]$ is a mapping defined as above, while $\mathcal{L}(\cdot)$ is simply regarded as a function, e.g., let $\mathcal{L}(p,q)=\left(m/2\right)p^2-\Phi_g(q)$, and $\mathcal{L}[x(t)]=\mathcal{L}(p,q)|_{p=\dot{x}(t),q=x(t)}$.

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